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Category: Limits

lim-x-pi-6-2ln-sin-x-ln-pi-sec-x-1-

Question Number 154719 by liberty last updated on 21/Sep/21 limxπ62ln[Γ(sinx)]lnπΓ(secx)1=? Answered by john_santu last updated on 21/Sep/21 $$\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{6}}} {\mathrm{lim}}\:\frac{\mathrm{2cos}\:{x}\:\Psi^{\mathrm{0}} \left(\mathrm{sin}\:\left({x}\right)\right)}{\mathrm{2tan}\:\left(\mathrm{2}{x}\right)\mathrm{sec}\:\left(\mathrm{2}{x}\right)\Gamma\left(\mathrm{sec}\:\left(\mathrm{2}{x}\right)\right)\Psi^{\mathrm{0}} \left(\mathrm{sec}\:\left(\mathrm{2}{x}\right)\right)}= \